Explicit or general rule

The rule is a_n=n(n+2). Substituting n=1 gives 1(1+2)=3; n=2 gives 2(2+2)=8; n=3 gives 3(3+2)=15; and n=4 gives 4(4+2)=24. Therefore, the first four terms are (3, 8, 15, 24), so option C is correct. Option A begins with 2, but the correct first term for n=1 must be 3. Exam tip: For a general-term question, substitute n=1, 2, 3, and so on in order.

For the fifth term, substitute \(n=5\): \(a_5=5\cdot2^{5-1}=5\cdot2^4=5\cdot16=80\). Therefore, 80 is correct. The value 160 would result from incorrectly using \(2^5\) instead of \(2^{n-1}\). Exam tip: Read the exponent carefully before substituting the term number.

This is (3^2,4^2,5^2,6^2,\ldots), so (a_n=(n+2)^2). In square sequences relate the base number to (n).

The sign alternates and the magnitude is (n), so (a_n=(-1)^{n+1}n). In alternating signs choose the power of ((-1)^n) carefully.

Substitute n=4 into both rules. \(a_4=2(4)+1=9\) and \(b_4=4^2-1=15\). Therefore, \(a_4+b_4=9+15=24\). A result such as 26 comes from evaluating one of the terms incorrectly. Exam tip: find each sequence term separately before adding them.

The direct answer is D: 89. The terms are 5, 11, 19, 29. Their differences are 6, 8, and 10, so the next differences are 12, 14, 16, and 18. Continuing gives 41, 55, 71, and 89 as the fifth through eighth terms. An explicit rule also confirms this: the rule is sufficiently represented by the supplied expression \\(a_n=n^2+3n+1\\); at n=8, \\(a_8=8^2+3(8)+1=64+24+1=89\\). Option D is correct. Option A, 77, does not result from the next-difference pattern and is not the eighth value. Option B, 81, and option C, 85, are also not obtained by continuing the even-number difference pattern. A useful check is to substitute n=1: \\(1+3+1=5\\), and n=2: \\(4+6+1=11\\), so the rule matches the start. Memory cue: when differences increase by 2, continue those differences carefully; do not assume the sequence has a constant difference.

Taking the term number as \(n=1,2,3,\ldots\), the rule \(a_n=n^2+2n-1\) gives \(a_1=2\), \(a_2=7\), \(a_3=14\), and \(a_4=23\). Hence, option A is correct. The closest distractor, \(n^2+n\), gives the second term as \(6\), not \(7\). Exam tip: when first differences increase regularly, such as \(5,7,9\), test a quadratic rule.

The general term is (a_n=4n+4), and (4n+4=50) gives (n=\frac{23}{2}). If the position is not a natural number, the given number is not a term.

These are squares of odd numbers (1^2,3^2,5^2,7^2,\ldots), so (a_n=(2n-1)^2). In squares observe the base pattern.

The rule for the terms is \(a_n=5^n\). Substituting \(n=4\), we get \(a_4=5^4=5\times5\times5\times5=625\). Option 125 is the value of \(5^3\), so it is a close but incorrect distractor. Exam tip: in an explicit rule, substitute the required term number directly for \(n\).

The numerator is (2n-1) and the denominator is (n+1), so (a_n=\frac{2n-1}{n+1}). In a fraction sequence form rules for both parts separately.

(a_5-a_2=9) and there are three gaps, so (d=3), then (a_n=3n+1). From two given terms first find the common difference.

Substitute \(n=1,2,3\) into the rule: \(a_1=1^2+3(1)-1=3\), \(a_2=2^2+3(2)-1=9\), and \(a_3=3^2+3(3)-1=17\). Thus, the first three terms are \((3, 9, 17)\), so option B is correct. In option A, the second term is 9, but the first and third terms do not follow the rule. Exam tip: for a general-term question, begin by substituting \(n=1\).

Given \(a_n=kn+2\). Substituting \(n=7\) gives \(a_7=7k+2\). Thus, \(7k+2=37\), so \(7k=35\) and \(k=5\). If \(k=6\), then \(a_7=44\), not 37. Exam tip: substitute the given term number into the general term to find an unknown constant.

The direct answer is option A, \(a_n=6n-2\). Each next term is 6 greater, so this is an arithmetic sequence with first term \(a_1=4\) and common difference \(d=6\). Use \(a_n=a_1+(n-1)d\): \(a_n=4+(n-1)6=4+6n-6=6n-2\). Check the first terms: \(n=1\) gives 4, \(n=2\) gives 10, \(n=3\) gives 16, and \(n=4\) gives 22. Option A therefore matches both the starting value and the repeated increase. Option B gives 10 at \(n=1\), so it ignores the stated first term. Option C gives 10 at \(n=1\) and increases by 6, but starts incorrectly. Option D gives 2 at \(n=1\), so its starting value is also wrong. A common mistake is writing \(a_1+nd\); the first gap is used zero times at the first term, hence \(n-1\).

The correct rule is \(a_n=n^2+n+1\). Substituting \(n=1,2,3,4\) gives \(3,7,13,21\), respectively. In option A, the second term is \(6\), so it does not match the sequence. In an exam, verify a rule using at least the first two or three terms.

The general rule gives the value of the term directly when its position is known: \\(a_n=\\frac{n}{n+2}\\). We need to find the position n for which the term equals \\(\\frac{5}{7}\\). Thus, substitute the required value and solve the resulting equation. Since a term number is normally a positive integer, the valid solution must be checked as a whole-number position.

Set \\(\\frac{n}{n+2}=\\frac{5}{7}\\). Cross-multiplication gives \\(7n=5(n+2)\\), so \\(7n=5n+10\\). Subtracting \\(5n\\) from both sides gives \\(2n=10\\), hence \\(n=5\\). Checking gives \\(a_5=\\frac{5}{5+2}=\\frac57\\). Therefore, the fifth term, option D, is correct.

Given \(a_n=2n^2+3\), \(a_3=2(3)^2+3=18+3=21\) and \(a_4=2(4)^2+3=32+3=35\). Therefore, \(a_3+a_4=21+35=56\). The value 52 may result from not including the constant term \(+3\) correctly in both terms. Exam tip: substitute the value of \(n\) and calculate each term separately before adding.

Taking the term number as \(n=1,2,3,4,\ldots\), the rule \(a_n=n^2-1\) gives \(1^2-1=0\), \(2^2-1=3\), \(3^2-1=8\), and \(4^2-1=15\). Hence, option B is correct. Option C gives \(0,2,4,6,\ldots\), which is a linear pattern and does not match the given sequence. Exam tip: substitute \(n=1,2,3,4\) to verify an explicit rule quickly.

Substitute \(n=1,2,3,4\) respectively: \(a_1=2^1+1=3\), \(a_2=2^2+2=6\), \(a_3=2^3+3=11\), and \(a_4=2^4+4=20\). Hence, the correct sequence is \((3,6,11,20)\). Option C has the correct first and fourth terms, but its second and third calculations are incorrect. Exam tip: evaluate \(2^n\) first, then add \(n\).

The magnitude is (3,5,7,9,\ldots) and the signs start negative and alternate, so (a_n=(-1)^n(2n+1)). In alternating signs always check the sign of the first term.

Given \(n^2+2n=80\), we get \(n^2+2n-80=0\). Factoring gives \((n+10)(n-8)=0\), so \(n=-10\) or \(n=8\). Since a sequence index \(n\) is a positive integer, \(n=8\) is valid. Option 9 is a close distractor, but \(a_9=81+18=99\), not 80. Exam tip: for term-position questions, check whether a negative root is allowed before selecting it.

For \(a_n=n^2-2n+2\), substituting \(n=1,2,3,4\) gives \(1,2,5,10\), respectively. Hence, it is the required general term. Option B gives the first term as \(2\), so it cannot represent the sequence. In an exam, verify a proposed general term by substituting the first few values of \(n\).

The correct rule is \(a_n=n^2+5n\). Substituting \(n=1,2,3,4\) gives \(6,14,24,36\), respectively. Also, the first differences are \(8,10,12\), whose differences are \(2\); hence the sequence should have a quadratic rule. Option B gives the first term as 6, but for \(n=2\) it gives 12, not 14. Exam tip: test a proposed rule with at least the first three values, \(n=1,2,3\).

In a linear rule the coefficient of (n) is (-3), so the common difference is (-3). Read the coefficient of (a_n) carefully.